Let f(n) = n^3-3n^2+2n-990f(n)=n3−3n2+2n−990
By the rational roots theorem, any rational roots of f(n) = 0f(n)=0 must be of the form p/qpq where pp and qq are integers, q != 0q≠0, pp a divisor of the constant term -990−990 and qq a divisor of the coefficient, 11, of the term n^3n3 of highest degree.
So the only possible rational roots are the factors of 990990, viz
+-1±1, +-2±2, +-3±3, +-5±5, +-6±6, +-9±9, +-10±10, +-11±11, +-18±18, +-22±22, +-30±30, +-33±33, +-45±45, +-55±55, +-90±90, +-99±99, +-110±110, +-165±165, +-198±198, +-330±330, +-495±495, +-990±990
That's rather a lot of factors to check (and I'm not sure I found all of them), so let's try to get closer:
n^3-3n^2+2n = 990n3−3n2+2n=990
So try n^3 ~= 990n3≅990, say n ~= 10n≅10
f(10) = 1000-300+20-990 = -270f(10)=1000−300+20−990=−270
f(11) = 1331-363+22-990 = 0f(11)=1331−363+22−990=0
So (n-11)(n−11) is a factor.
Divide f(n)f(n) by (n-11)(n−11) to find:
f(n) = n^3-3n^2+2n-990 = (n-11)(n^2+8n+90)f(n)=n3−3n2+2n−990=(n−11)(n2+8n+90)
The discriminant of n^2+8n+90n2+8n+90 is:
8^2-(4*1*90) = 64 - 360 = -29682−(4⋅1⋅90)=64−360=−296
So there are no more linear factors with real coefficients.
